Timeline for Representing dimensions in Dirac delta function results
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 14, 2017 at 18:26 | comment | added | Ariana | Yeah I was just asking for notation. And yeah it shouldve been root not inverse root | |
| Jun 14, 2017 at 18:25 | comment | added | Wojciech Morawiec | @ArianaGrande The square root shouldn't be the inverse, because I'm talking about the units of $A$, which has inverse units of $\delta(p)$, which is the inverse of the inverse of $p$. Anyway, since you seem to be asking about notation, I'd just write something like $A = \frac{1}{\sqrt{2\pi\hbar}}\sqrt{\mathrm{kg}\mathrm{\frac{m}{s}}}$. | |
| Jun 14, 2017 at 18:20 | comment | added | Ariana | Yes, there is a problem there, the question is how should I express the extra units, oh and the $kg\frac{m}{s}$ should go inside the inverse root.(oh and the recent edits are just formatting changes, no content added) | |
| Jun 14, 2017 at 18:17 | comment | added | Wojciech Morawiec | @ArianaGrande Ok, I'm confused: In the edit to your question you say something along the lines of "ignoring unit consistency, the units do not match". What exactly is it you are asking? $A$ doesn't have dimensions of $(2\pi\hbar)^{-1/2}$, it has dimensions of $(2\pi\hbar)^{-1/2} (\mathrm{kg}\mathrm{\frac{m}{s}})^{1/2}$ because of the units of the delta distribution, which in this case has the dimension of inverse momentum. | |
| Jun 14, 2017 at 18:12 | comment | added | Ariana | However, in the example, $A^2$ is shown to be $\frac{1}{2\pi\hbar}$ for normalization purposes, so how we represent the units of the delta function in the original wave function $Ae^{\frac{\iota px}{\hbar}}$ | |
| Jun 14, 2017 at 18:08 | history | edited | Wojciech Morawiec | CC BY-SA 3.0 |
deleted 147 characters in body
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| Jun 14, 2017 at 18:06 | comment | added | Wojciech Morawiec | @ArianaGrande Nice, we are on the same page, then :) | |
| Jun 14, 2017 at 18:05 | comment | added | Ariana | Sorry, I've edited the error in the question, I meant for it to be $m^{-\frac{1}{2}}$ | |
| Jun 14, 2017 at 18:03 | history | answered | Wojciech Morawiec | CC BY-SA 3.0 |